Write the solution of set of


Write the solution of set of $\left|x+\frac{1}{x}\right|>2$.


We have:



CASE $1:$ When $x+\frac{1}{x}>0$, then $\left|x+\frac{1}{x}\right|=x+\frac{1}{x}$

Now, $\left|x+\frac{1}{x}\right|-2>0$

$\Rightarrow x+\frac{1}{x}-2>0$

$\Rightarrow \frac{x^{2}+1-2 x}{x}>0$

$\Rightarrow \frac{(x-1)^{2}}{x}>0$

$\Rightarrow x>0$ and $x \neq 1$

$\Rightarrow x \in(0,1) \mathrm{U}(1, \infty) \quad \ldots$ (i)

CASE $2:$ When $x+\frac{1}{x}<0$, then $\left|x+\frac{1}{x}\right|=-\left(x+\frac{1}{x}\right)$

Now, $\left|x+\frac{1}{x}\right|-2>0$


$\Rightarrow \frac{-x^{2}-1-2 x}{x}>0$

$\Rightarrow \frac{x^{2}+1+2 x}{x}<0 \quad$ [Multiplying both sides by $\left.-1\right]$

$\Rightarrow \frac{(x+1)^{2}}{x}<0$

$\Rightarrow x<0$ and $x \neq-1$

$\Rightarrow x \in(-\infty,-1) \mathrm{U}(-1,0) \quad \ldots$(ii)

Thus, the solution set of the given inequation is the union of (i) and (ii)

$(0,1) \mathrm{U}(1, \infty) \cup(-\infty,-1) \mathrm{U}(-1,0)=R-\{-1,0,1\}$

$\therefore x \in R-\{-1,0,$,


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